If a -k- b -2 -3 k- and c -2 -k- find a unit vector parallel to the vector 2 a - b -3 c-

We have, a=î+ĵ+k̂, b=2î ĵ+3k̂ and c=î 2ĵ+k̂ Let v=2a b+3c=2(î+ĵ+k̂) (2î ĵ+3k̂)+3(î 2ĵ+k̂)=3î 3ĵ+2k̂ Now, |v|=|3î 3ĵ+2k̂|=√(3^2+( 3)^2+2^2)=√(22) Hence, the ...

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Byju's Answer

Standard XII

Mathematics

Scalar Triple Product

If a =î+ĵ+k̂,...

Question

If $a =^i +^j +^k, b = 2^i -^j + 3^k a n d c =^i - 2^j +^k$ find a unit vector parallel to the vector 2a - b + 3c.

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Solution

We have,
$a =^i +^j +^k, b = 2^i -^j + 3^k a n d c =^i - 2^j +^k$
Let $v = 2 a - b + 3 c = 2 (^i +^j +^k) - (2^i -^j + 3^k) + 3 (^i - 2^j +^k) = 3^i - 3^j + 2^k$
Now, $| v | = | 3^i - 3^j + 2^k | = \sqrt{3^{2} + (- 3)^{2} + 2^{2}} = \sqrt{22}$
Hence, the unit vector along v is
$^v = \frac{v}{| v |} = \frac{3^i - 3^j + 2^k}{\sqrt{22}} = (\frac{3}{\sqrt{22}}^i - \frac{3}{\sqrt{22}}^j + \frac{2}{\sqrt{22}}^k)$

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If a=^i+^j+^k, b=2^i−^j+3^k and c=^i−2^j+^k find a unit vector parallel to the vector 2a - b + 3c.

We have, a=^i+^j+^k, b=2^i−^j+3^k and c=^i−2^j+^k Let v=2a−b+3c=2(^i+^j+^k)−(2^i−^j+3^k)+3(^i−2^j+^k)=3^i−3^j+2^k Now, |v|=|3^i−3^j+2^k|=√32+(−3)2+22=√22 Hence, the unit vector along v is ^v=v|v|=3^i−3^j+2^k√22=(3√22^i−3√22^j+2√22^k)

If $a =^i +^j +^k, b = 2^i -^j + 3^k a n d c =^i - 2^j +^k$ find a unit vector parallel to the vector 2a - b + 3c.